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Specifically, if a circuit can amplify a 10mV (10kHz) injected sinusoidal signal to about 6V, why does it not amplify the 10kHz RF output of a L-C tank circuit?

The tank circuit below has a resonant frequency of 10kHz.

Injected Signal. To check the amplifier works, I removed the tank circuit (ie C1 and L1) and used an oscilloscope to inject a sinusoid (20mV peak to peak, 10kHz) into C4. At the time this test was done, R1=R2=47k.

At A, I got 56mV peak-to-peak. At B, I got 5.7V peak-to-peak (output saturated, so it looks like a square wave). (With no injected signal, A=14mV p2p, B=90mV p2p).

Tank Circuit Signal. Using the tank circuit (C1 and L1), C1 is a variable capacitor to do fine tuning. The entire circuit was made on a breadboard.

The transmitter is an NE555 square wave generator module with its output connected to 30 feet of speaker wire which goes out the window in a straight horizontal line. The receiver (tank circuit) is about 1 foot away from the transmitter for the readings below.

With R1=R2=47k (ie the same values as for the injected signal), the outputs at A and B were very small (B=300mV p2p).

I then set the values of R1=R2=1M to increase gain. Point B then had an output of B=700mV p2p.

For comparison, with transmitter OFF, B=130mV p2p.

So, here is my question (repeated): If a circuit can amplify a 10mV injected signal to about 6V, why does it not amplify the output of a L-C tank ciruit?

If I isolate the tank circuit then its output is about 200mV p2p. I understand this 200mV is reduced if the circuit is loaded on its output by a low impedance sink, which is why I chose the high impedance TL084 IC as an amplifier.

I have tried many different amplifier circuits before this, and I get the same sort of result as above every time.

Please feel free to interrogate me and this circuit for any questions you may have.

Context: My cat was run over last year (he survived, but now has a metal bar through his hips). My objective is to get this cat-fence working before he dies of old age.

BTW R5=680k.

Note1: The TL084 has a voltage divider to set the non-inverting inputs to 3V.

Note2: The RF choke is used in place of a ferrite rod to reduce weight (see Note3).

Note3: The low RF frequency of 10kHz is used because the intention is for the transmitter to be part of a cat-fence (the transmitter antenna will be a long wire which goes around my garden). The cat will wear the receiver below on his collar (the 18cm aerial will be on his collar). When the signal strength is large, this will set off a buzzer on his collar, to discourage him fro leaving the garden. (If I can make the cat collar circuit + buzzer + battery large enough then just the weight of it should stop the cat from leaving the garden).

Note4: The voltage divider R6/R7 is implemented as a potentiometer, with the non-inverting inputs to TL084 connected to the middle pin of the potentiometer.

Note5: The diode D1 is not really relevant yet, but it is there to rectify the output so I can measure signal strength as a DC signal.

Note6: The power supply is a 9V battery.

TL084 RF amplifier circuit

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    \$\begingroup\$ "...which is why I chose the high impedance TL084 IC as an amplifier." Which you completely negated by biasing it with a source that has an impedance of about 10 kΩ (R6 || R7). This application is all about the extremely high source impedance of your input signal. Any loading at all will reduce the voltage to nothing. For starters, try applying the bias at the other end of L1 and eliminate C4. Instead, put a larger capacitor between L1 and ground. \$\endgroup\$
    – Dave Tweed
    Commented Feb 3 at 14:18
  • \$\begingroup\$ Point taken, thanks. I have made the changes you suggest in Figure 3 at the bottom of my post - can you confirm that Figure 3 is what you intended? \$\endgroup\$ Commented Feb 3 at 15:39
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    \$\begingroup\$ Yes, that's the idea. Now, do you have any way to measure how much signal you're actually getting at the antenna connection? \$\endgroup\$
    – Dave Tweed
    Commented Feb 3 at 17:30
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    \$\begingroup\$ @JamesVStone - Hi, We've been alerted to the large number of edits you've been making to the question. It seems like you are documenting a project, which is not the purpose of Stack Exchange. || We expect a question to be (ideally) a single clear question, with all details given when first asked. Some edits might be needed, but only to support the original question. After the 20+ edits, and what appears to be obsolete information left in the text, I'm not clear what the question now is. || What's going on? What is the specific question? Why are edits still needed and when will they stop? TY \$\endgroup\$
    – SamGibson
    Commented Feb 5 at 14:57
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    \$\begingroup\$ I have rolled this back to revision 4. Any further updates invalidates answers so please don't do that. I did warn you about this in the comment under my answer: I'm sorry but your question amendment invalidated my answer. It boils down to accepting responsibility for posting correct information and, when someone makes an answer (in good faith) based on incorrectly posted material then the question is rolled-back to how it was when the answer was made. \$\endgroup\$
    – Andy aka
    Commented Feb 7 at 14:07

2 Answers 2

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TL084 has gain bandwidth product around 3 MHz,
That makes available voltage gain at 10kHz a few hundred.

If a circuit can amplify a 10mV injected signal to about 6V, why does it not amplify the output of a L-C tank ciruit?

Most problematic is how the injected signal is introduced to the tank. An 18cm antenna @ 10kHz can only inject much signal if the tank is very, very high impedance. That's because such an antenna has a tiny equivalent capacitance (to ground, or cat).

Tank impedance is limited by inductor equivalent resistance. Searching for 100mH chokes suggests a Q of 100 might be achieved, with series resistance of 82 ohms. That makes parallel equivalent resistance 820kohms. A fraction of a picofarad antenna won't inject much voltage into that impedance.
Furthermore, breadboard capacitance will seriously deteriorate performance at such high impedances as this.

This kind of antenna is often called voltage-probe antenna, and requires an extremely high-impedance preamp. Often, a J-FET common-drain buffer is used to transform the extremely-high antenna impedance down to a lower impedance that is easier to amplify.

For your TL084, you might try connecting antenna direct to "+" input. For safety, a very high bias resistor might help snub electrostatic spikes. On a breadboard, the "+" input and its bias resistor should be isolated. Nevertheless, package capacitance and TL084 input capacitance will likely seriously load that short antenna.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thanks. I am reading up on voltage-probe antenna now. Interesting design. It looks like the output should be 10kHz. I may experiment with a longer aerial (so I may have to get a bigger cat to fit the bigger collar). More anon ... \$\endgroup\$ Commented Feb 3 at 18:13
  • \$\begingroup\$ @DaveTweed makes another different suggestion, whose circuit #3 you got right. His suggestion has no gain at DC, whereas mine has significant gain at DC - which is a potential problem with ESD as your cat romps, flinging antenna about. However, mine likely has more gain because its input impedance is higher (if your construction is done carefully). Good luck herding your cat. \$\endgroup\$
    – glen_geek
    Commented Feb 3 at 18:28
  • \$\begingroup\$ Glen, Many thanks. Using R2=1k just sets the circuit oscillating at resonant frequency. But if R2=2k then all is well. Using R1=4M (I only hav 4 1M resistors atm), I get 2.7V p2p, even if transmitter is 12" away, I get 1V p2p. With transmitter off, I get 133mV p2p. Can I just add this to my original design (ie boost the output using a second amplifier within TL084)? I should have asked you guys months ago; thank you so very much. I'll post the final circuit when it is working. \$\endgroup\$ Commented Feb 3 at 19:58
  • \$\begingroup\$ The answer to the queston: Can I add this to my circuit (ie boost the output using a second amplifier within TL084)? Is "yes", according to my new 2-stage circuit readings. I'll post a circuit diagram of this 2-stage amplifying circuit tomorrow, with results. \$\endgroup\$ Commented Feb 3 at 22:13
  • \$\begingroup\$ Glen, I had to look up ESD (electrostatic discharge). I plan to rectify the output and compare the DC output with a threshold (using LM393?), so I hope transient ESDs will not be a problem. But thanks for the warning. \$\endgroup\$ Commented Feb 3 at 22:16
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Your design has an error.

You are using the left op-amp to amplify the tank AC signal by about 100 then, you are feeding that signal to a 2nd op-amp on the inverting input but, that 2nd op-amp does not have the feedback connected up at all properly. The junction of R2 and R4 do not connect to the inverting pin so, this circuit just will not work.

enter image description here

I would also recommend a slight modification to your op-amp bias circuit (an added 100 kΩ resistor and a circa 10 μF capacitor: -

enter image description here

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  • \$\begingroup\$ Thank you. That was an error in my drawing, not in the circuit. I have uploaded a corrected diagram. Can I ask why you advise adding the 100k resistor and the 10 microF capacitor to the voltage divider? \$\endgroup\$ Commented Feb 3 at 13:13
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    \$\begingroup\$ @JamesVStone I'm sorry but your question amendment invalidated my answer. It boils down to accepting responsibility for posting correct information and, when someone makes an answer (in good faith) based on incorrectly posted material then the question is rolled-back to how it was when the answer was made. \$\endgroup\$
    – Andy aka
    Commented Feb 3 at 13:18
  • \$\begingroup\$ 100k is a bit small, limiting the Q of the tank to 16. \$\endgroup\$
    – John Doty
    Commented Feb 3 at 13:37
  • \$\begingroup\$ Andy, I left my original drawing with its error in my posted question to leave a transparent trail for others. There was not intention to invalidate your comment, which was most welcome. \$\endgroup\$ Commented Feb 3 at 14:25
  • \$\begingroup\$ John and Andy, Thanks, I just realised the tank circuit is loaded by the (low impedance) voltage divider, which is presumably why you suggested the 100k resistor. I'll see if that helps. \$\endgroup\$ Commented Feb 3 at 14:30

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